Maths

110 MATHEMATICS
i.e.,
910 = 91d
or, d =10
Therefore, a 20
= 10 + (20 – 1) × 10 = 200, i.e. 20th term is 200.
Example 13 : How many terms of the AP : 24, 21, 18, . . . must be taken so that their
sum is 78?
We know that S n
= ✁ ✂ ✄
Solution : Here, a = 24, d = 21 – 24 = –3, S n
= 78. We need to find n.
n
2
2 a ( n 1) d
n
n
☎ ✆
So, 78 = 48 ( ✝ ✞ ✞ n 1)( ✟ ✠
3) = 51✞
3n
2
2
or 3n 2 – 51n + 156 = 0
or n 2 – 17n + 52 = 0
or (n – 4)(n – 13) = 0
or
n =4or13
Both values of n are admissible. So, the number of terms is either 4 or 13.
Remarks:
1. In this case, the sum of the first 4 terms = the sum of the first 13 terms = 78.
2. Two answers are possible because the sum of the terms from 5th to 13th will be
zero. This is because a is positive and d is negative, so that some terms will be
positive and some others negative, and will cancel out each other.
Example 14 : Find the sum of :
Solution :
(i) the first 1000 positive integers
(ii) the first n positive integers
(i) Let S = 1 + 2 + 3 + . . . + 1000
n
Using the formula S n
= ( a ✝ l ) for the sum of the first n terms of an AP, we
2
have
S 1000
= 1000 (1 1000) ✝ = 500 × 1001 = 500500
2
So, the sum of the first 1000 positive integers is 500500.
(ii) Let S n
= 1 + 2 + 3 + . . . + n
Here a = 1 and the last term l is n.